Calculating the directional derivative of a function of two variables

In summary, the function z=f(x,y) has a directional derivative at (x,y)=(1,1) in the direction of the vector (1,2).
  • #1
wshfulthinker
8
0
Homework Statement

Consider the function:
z=f(x,y)= log(x^2 + y^2) (x,y)=/=(0,0)

Calculate the directional derivative of f(x,y) at (x,y)=(1,1) in the direction of the vector (1,2)


The attempt at a solution

When i tried to work out the unit vector from the point (1,1) to (1,2) i got (0,1).

I got partial derivative df/dx = 2x/(x^2 + y^2)
and partial derivative df/dy = 2y/(x^2 + y^2)

Then, for gradf at (1,1) i got (1,1)..

so, for directional derivative i got:

(gradf at (1,1)) x unit vector = (1,1).(0,1) = 1

But the answer is 3/root5

Does anyone know what i have done wrong? Thankyou
 
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  • #2
Welcome to PF!

Hi wshfulthinker! Welcome to PF! :smile:

(have a curly d: ∂ and a square-root; √ and a grad: ∇ and try using the X2 tag just above the Reply box :wink:)
wshfulthinker said:
Calculate the directional derivative of f(x,y) at (x,y)=(1,1) in the direction of the vector (1,2)

When i tried to work out the unit vector from the point (1,1) to (1,2) i got (0,1).

Nooo … you're msunderstanding "the direction of the vector (1,2)" …

it's not the point (1,2) (which is in direction (0,1) from (1,1), as you say) …

(1,2) is the actual direction that you're taking the derivative along.

(otherwise, your method is ok :wink:)
 
  • #3
Hi, thanks for the welcome and showing me the symbols!

I don't really get it though! where do i use the point (1,2). I'm not even sure what i worked out, i followed the method that were in my lecture notes which were worded almost the exact same way as my actual question (except it said find instead of calculate - i don't know if that means it's different)
 
  • #4
wshfulthinker said:
I don't really get it though! where do i use the point (1,2).

grrr! it's not a point! :rolleyes:

it's a direction … use it instead of your (0,1). :smile:
 
  • #5
Okay okay... so i am pretty crap with vectors! But i got the answer finally! :D I found a book which wrote the direction in the i + j form which made more sense to me and didn't make me think it was a point. And yes, i got the answer so i think i kind of understand it now... Thankyou! :)
 

1. How do you calculate the directional derivative of a function of two variables?

To calculate the directional derivative of a function of two variables, you can use the formula Df(x,y) = ∇f(x,y) · u, where ∇f(x,y) is the gradient of the function and u is the unit vector in the direction of interest.

2. What is the significance of the directional derivative in multivariable calculus?

The directional derivative measures the rate of change of a function in a specific direction, allowing us to understand how the function changes as we move in different directions from a given point. It is an important concept in multivariable calculus as it helps us analyze the behavior of functions in multiple dimensions.

3. How does the directional derivative relate to the partial derivatives of a function?

The directional derivative is related to the partial derivatives of a function through the gradient vector. The gradient vector contains the partial derivatives of a function with respect to each variable, and the directional derivative can be calculated using this vector and a unit vector in the desired direction.

4. Can the directional derivative be negative?

Yes, the directional derivative can be negative. This indicates that the function is decreasing in the direction of interest.

5. How can the directional derivative be used in real-world applications?

The directional derivative has many real-world applications, such as in physics, engineering, and economics. For example, it can be used to calculate the rate of change of temperature in different directions in a given space, or to analyze the sensitivity of a system to changes in different variables.

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